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CoCl(4)^(2-)(aq) is blue in color while ...

`CoCl_(4)^(2-)(aq)` is blue in color while `[Co(H_(2)O)_(6)]^(2+)` is pink. The color of reaction mixture
`Co(H_(2)O)_(6)^(2+)(aq)+4Cl^(-)(aq)hArr CoCl_(4)^(2-)(aq)+6H_(2)O(l)`
is blue at room temprature while it is pink when cooled heance

A

reaction is exothermic

B

reaction is endothermic

C

equlibrium will shift in forward direction on adding water to reaction mixture

D

equilibrium will not shift on adding water to equilibrium mixture.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the reaction and the colors of the complexes involved. The reaction is: \[ \text{Co(H}_2\text{O)}_6^{2+}(aq) + 4\text{Cl}^-(aq) \rightleftharpoons \text{CoCl}_4^{2-}(aq) + 6\text{H}_2\text{O}(l) \] 1. **Identify the colors of the complexes**: - The complex \(\text{Co(H}_2\text{O)}_6^{2+}\) is pink. - The complex \(\text{CoCl}_4^{2-}\) is blue. 2. **Determine the color of the reaction mixture at different temperatures**: - At room temperature, the reaction mixture is blue, indicating that the equilibrium favors the formation of \(\text{CoCl}_4^{2-}\). - When cooled, the color changes to pink, indicating that the equilibrium shifts towards the formation of \(\text{Co(H}_2\text{O)}_6^{2+}\). 3. **Analyze the effect of temperature on the reaction**: - Since the color changes from blue to pink upon cooling, we can conclude that the forward reaction (formation of \(\text{CoCl}_4^{2-}\)) is endothermic. This means that heat is absorbed during this reaction. - According to Le Chatelier's principle, if the temperature decreases, the equilibrium will shift in the direction that produces heat, which is the exothermic direction (backward in this case). 4. **Effect of adding water**: - If water is added to the system, the concentration of all species will decrease. This will also shift the equilibrium towards the side with more moles of solute, which is the backward direction (towards \(\text{Co(H}_2\text{O)}_6^{2+}\)). 5. **Conclusion**: - The correct statement is that the reaction is endothermic, and the equilibrium shifts to the left (backward direction) upon cooling or adding water. ### Summary of Steps: 1. Identify colors of the complexes. 2. Determine the color of the reaction mixture at different temperatures. 3. Analyze the effect of temperature on the reaction. 4. Consider the effect of adding water on the equilibrium. 5. Conclude the nature of the reaction and the direction of the equilibrium shift.

To solve the problem, we need to analyze the reaction and the colors of the complexes involved. The reaction is: \[ \text{Co(H}_2\text{O)}_6^{2+}(aq) + 4\text{Cl}^-(aq) \rightleftharpoons \text{CoCl}_4^{2-}(aq) + 6\text{H}_2\text{O}(l) \] 1. **Identify the colors of the complexes**: - The complex \(\text{Co(H}_2\text{O)}_6^{2+}\) is pink. - The complex \(\text{CoCl}_4^{2-}\) is blue. ...
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Assertion (A) : On cooling in a freezing mixture, colout of the mixture turns to pink from deep blue for a reaction. underset(("Pink"))(Co(H_(2)O)_(6)^(2+))(aq)+4CI^(Theta)hArr underset((Blue))(CoCI_(4)^(2-)(aq))+6H_(2)O(l) Reason (R) : The reaction is endothermic in forward reaction,so on cooling the reaction, deep blue colour appears.

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